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Lesson Explainer: Linear Momentum Mathematics

In this explainer, we will learn how to calculate the momentum of a particle moving in a straight line using the formula 𝐻=π‘šπ‘£.

Imagine two objects: a truck moving at 30 mph along a road and a paper airplane moving at 2 mph through the air. Which object would require a greater force to stop it in the same amount of time? Intuitively, we know that the truck would require the greater force to stop it because it has a greater mass and it is moving faster. We can say that the truck has a greater momentum. Momentum can be thought of as a measure of how difficult it is to stop an object that is moving.

The two factors that contribute to an object’s momentum are its mass, π‘š, and its velocity, ⃑𝑣. The greater the mass of the object, the greater its momentum, and similarly the greater the velocity of the object, the greater its momentum. Knowing this, we can now define momentum mathematically.

Definition: Momentum

The momentum of an object, ⃑𝐻, is equal to its mass, π‘š, multiplied by its velocity, ⃑𝑣: ⃑𝐻=π‘šβƒ‘π‘£.

Since velocity is a vector quantity and mass is a scalar quantity, momentum is a vector quantity. However, often we just want the magnitude of the momentum, which we can write as ‖‖⃑𝐻‖‖=β€–β€–π‘šβƒ‘π‘£β€–β€–.

Since mass is a scalar quantity, we can take it outside of the magnitude sign, giving us ‖‖⃑𝐻‖‖=π‘šβ€–β€–βƒ‘π‘£β€–β€–.

On the right-hand side, ‖‖⃑𝑣‖‖ is just the magnitude of the velocity, or the speed. We can denote the magnitude of the momentum as just 𝐻 and the speed as just 𝑣, giving us 𝐻=π‘šπ‘£.

This is the scalar form of the definition of momentum, and this is what we will end up using most of the time.

So, for example, imagine a bowling ball with a mass of 12 kg moving at a speed of 5 m/s along the lane of a bowling alley. What is the momentum of the bowling ball? We can substitute these values into 𝐻=π‘šπ‘£ to get 𝐻=12Γ—5/kgms, giving us 60 kgβ‹…m/s.

This shows us that the standard unit of momentum is kgβ‹…m/s, or β€œkilogram-metres per second,” but momentum can be measured in other units; in fact, any unit that is a unit of mass multiplied by a unit of speed is a valid unit of momentum.

Let’s have a look at some example questions.

Example 1: Finding the Momentum of an Object given Its Speed

Determine the momentum of a car of mass 2.1 tonnes moving at 42 km/h.

Answer

The mass of the car is 2.1 tonnes, and the speed of the car is 42 km/h. These are the two factors that affect an object’s momentum, and we can substitute them straight into the formula 𝐻=π‘šπ‘£ to find the momentum: 𝐻=π‘šπ‘£π»=2.1Γ—42/𝐻=88.2β‹…/tkmhtkmh

Here, we have used the symbol for the tonne, which is t. This answer is in units of tonne-kilometres per hour, which is a slightly unusual unit, but still a valid unit of momentum.

Example 2: Finding the Momentum of a Falling Body after a Given Distance

Calculate the momentum of a stone of mass 520 g after it has fallen 8.1 m vertically downward. Consider the acceleration due to gravity to be 𝑔=9.8/ms.

Answer

In this question, we are not told what the speed of the stone is, which we need to know in order to calculate the momentum, so we are going to have to work that out first.

We are told how far the stone falls, and what its acceleration is as it falls. The question says nothing about the initial motion of the stone, so we can assume that initially the stone is not moving. Knowing these three pieces of information, there is a formula that we can use to find the speed of the particle after it has fallen 8.1 mβ€”it is one of the kinematic equations: 𝑣=𝑒+2π‘Žπ‘ , where 𝑒 is the initial speed of the stone, π‘Ž is the acceleration of the stone, 𝑠 is the distance that the stone moves, and 𝑣 is the final speed of the stone. Since 𝑒 is zero, π‘’οŠ¨ is also zero, and this formula simplifies to just 𝑣=2π‘Žπ‘ .

We are trying to find 𝑣, so we want to make 𝑣 the subject of the equation. To do this, we need to square root each side of the equation: βˆšπ‘£=√2π‘Žπ‘ π‘£=√2π‘Žπ‘ .

We can now substitute in the values we have for π‘Ž and 𝑠: 𝑣=√2Γ—9.8/Γ—8.1𝑣=√158.76/𝑣=12.6/.msmmsms

So the stone starts at rest; it then begins to fall at 9.8 m/s2 due to gravity, and by the time it has fallen a distance of 8.1 m, it is moving at a speed of 12.6 m/s.

The other quantity that we need to know in order to calculate the momentum of the stone is its mass, which we were given. The mass was given in grams; let’s convert that to the standard unit of kilograms. There are 1β€Žβ€‰β€Ž000 g in 1 kg, so 520 g = 0.52 kg.

We can now substitute the values for the mass and speed of the stone into the formula for momentum: 𝐻=π‘šπ‘£π»=0.52Γ—12.6/𝐻=6.552β‹…/.kgmskgms

So by the time the stone had fallen a distance of 8.1 m, it had a momentum of 6.552 kgβ‹…m/s.

Example 3: Finding the Increase in Momentum of a Body Moving with Uniform Acceleration after a Given Time

A body of mass 17 kg moves in a straight line with constant acceleration of 1.8 m/s2. Its initial velocity is 22.3 m/s. Find the increase in its momentum in the first 5 seconds.

Answer

This question asks us to find the increase in momentumβ€”the change in momentumβ€”over a given time period. In order to do that, we need to work out the difference between its final momentum and its initial momentum. We can express this mathematically as Δ𝐻=π»βˆ’π»=π‘šπ‘£βˆ’π‘šπ‘£, where Δ𝐻 is the change in momentum, 𝐻 is the final momentum, 𝐻 is the initial momentum, π‘£οŠ¨ is the final speed, π‘£οŠ§ is the initial speed, and π‘š is the mass.

We already know the initial speed of the object and its mass; the only thing we do not know is its final speed, so we will have to work that out first.

Since the object undergoes constant acceleration, we can use one of the kinematic equations to find the final speed of the object: 𝑣=𝑣+π‘Žπ‘‘, where π‘Ž is the acceleration of the object and 𝑑 is how long the object accelerates for. We know all three values that are used on the right-hand side of this equation, so we can use it to find π‘£οŠ¨, the final speed: 𝑣=22.3/+1.8/Γ—5𝑣=31.3/.msmssms

So we now know the initial speed, the final speed, and the mass of the object. We can substitute these values into our equation from earlier to find the change in momentum: Δ𝐻=π»βˆ’π»Ξ”π»=π‘šπ‘£βˆ’π‘šπ‘£Ξ”π»=17Γ—31.3/βˆ’17Γ—22.3/Δ𝐻=532.1β‹…/βˆ’379.1β‹…/Δ𝐻=153β‹…/.kgmskgmskgmskgmskgms

So, in the first 5 seconds of its motion, the momentum of the object increases by 153 kgβ‹…m/s.

Example 4: Finding the Momentum of an Object at a Given Time given Its Displacement with respect to Time

A car of mass 1β€Žβ€‰β€Ž350 kg moves in a straight line such that at time 𝑑 seconds its displacement from a fixed point on the line is given by 𝑠=ο€Ή6π‘‘βˆ’3𝑑+4ο…οŠ¨m. Find the magnitude of the car’s momentum at 𝑑=3s.

Answer

In this question, we are given a function for the position of the car that depends only on time. In order to find the momentum of the car at a particular time, we are going to need to know its velocity at that time. In order to get the velocity at a particular time, we are going to need to get a general function for the velocity of the car over time.

Recall that the velocity of an object, ⃑𝑣, is defined as the rate of change of the displacement of the object. It is the derivative of the displacement of the object with respect to time: ⃑𝑣=⃑𝑠𝑑.dd

Since this question is just about an object moving in one dimension, we can just use scalar quantities to represent the velocity, 𝑣, and the displacement, 𝑠, giving us 𝑣=𝑠𝑑.dd

Taking the derivative with respect to 𝑑 of the function for 𝑠 that we were given in the question, we get 𝑣=𝑑6π‘‘βˆ’3𝑑+4𝑣=(12π‘‘βˆ’3)/.ddms

We want to know the speed of the car at 𝑑=3s, so let’s substitute this value for the time into the above equation: 𝑣=(12Γ—3βˆ’3)/𝑣=33/.msms

We now know both the speed of the car and the mass of the car, so we can substitute these values into the formula for the momentum of the car: 𝐻=π‘šπ‘£π»=1350Γ—33/𝐻=44550β‹…/.kgmskgms

At 𝑑=3s, the car has a momentum of 44β€Žβ€‰β€Ž550 kgβ‹…m/s.

Example 5: Finding the Momentum of a Body at a Given Time given an Expression of Its Position with Time

A body of mass 7 kg is moving in a straight line. Its position vector at a time 𝑑 is given by the relation βƒ‘π‘Ÿ(𝑑)=𝑑+5⃑𝑖+𝑑+π‘‘ο…βƒ‘π‘—οŠ¨οŠ©, where β€–β€–βƒ‘π‘Ÿβ€–β€– is measured in metres and 𝑑 in seconds. Determine the body’s momentum after 2 seconds.

Answer

In this question, we are given a function for the position of the object that depends only on time. In order to find the momentum of the object at a particular time, we are going to need to know the velocity of the object at that time. In order to get the velocity at a particular time, we are going to need to get a general function for the velocity of the object over time.

The velocity, ⃑𝑣, of an object is the derivative of its displacement, βƒ‘π‘Ÿ, with respect to time: ⃑𝑣=βƒ‘π‘Ÿπ‘‘βƒ‘π‘£=𝑑𝑑+5⃑𝑖+𝑑+𝑑⃑𝑗⃑𝑣=𝑑𝑑+5⃑𝑖+𝑑𝑑+𝑑⃑𝑗⃑𝑣=(2𝑑)⃑𝑖+ο€Ή3𝑑+1⃑𝑗.dddddddd

The momentum, ⃑𝐻, of an object is equal to its mass, π‘š, times its velocity: ⃑𝐻=π‘šβƒ‘π‘£βƒ‘π»=π‘šο€Ί(2𝑑)⃑𝑖+ο€Ή3𝑑+1⃑𝑗⃑𝐻=π‘š(2𝑑)⃑𝑖+π‘šο€Ή3𝑑+1⃑𝑗.

We now have a formula that will give us the vector momentum of the object at a time 𝑑. We can substitute the values for the mass of the object and the time into the equation: ⃑𝐻=7Γ—(2Γ—2)⃑𝑖+7Γ—ο€Ή3Γ—2+1⃑𝑗⃑𝐻=ο€Ί28⃑𝑖+91⃑𝑗⋅/.kgms

The momentum of the object after 2 seconds is ο€Ί28⃑𝑖+91⃑𝑗 kgβ‹…m/s.

At the start of this explainer, we wrote a mathematical definition of momentum: ⃑𝐻=π‘šβƒ‘π‘£.

If we take the derivative with respect to time of each side of the equation, dddddddddd𝑑⃑𝐻=π‘‘ο€Ήπ‘šβƒ‘π‘£ο…βƒ‘π»π‘‘=π‘šβƒ‘π‘£π‘‘βƒ‘π»π‘‘=π‘šβƒ‘π‘Ž, we can see that the derivative of the momentum with respect to time is equal to the mass of the object, π‘š, times the acceleration of the object, βƒ‘π‘Ž. But π‘šβƒ‘π‘Ž is also equal to the force on the object, ⃑𝐹, so ⃑𝐹=⃑𝐻𝑑.dd

In the case where the acceleration of an object is constant or we know the average acceleration, we can find the change in momentum, Δ⃑𝐻, of an object if we know its mass, its acceleration, and the time for which it accelerates, Δ𝑑: Δ⃑𝐻=π‘šβƒ‘π‘ŽΞ”π‘‘ or, in scalar form, Δ𝐻=π‘šπ‘ŽΞ”π‘‘.

This formula only works if our acceleration, π‘Ž, is constant. In the case where acceleration changes as time, 𝑑, changes, we can use the equation we found earlier for the rate of change of momentum with respect to time, dd⃑𝐻𝑑=π‘šβƒ‘π‘Ž, and integrate both sides of this formula with respect to 𝑑. Doing this will give us ⃑𝐻𝑑𝑑=ο„Έπ‘šβƒ‘π‘Žπ‘‘Ξ”βƒ‘π»=π‘šο„Έβƒ‘π‘Žπ‘‘.ddddd

On the left, we can see that the integral removes the change in time and all we are left with is a change in momentum. On the right, we can take π‘š out of the integral since it is a constant. Let us now consider what this formula would look like if we were trying to find the change in momentum over the time interval [𝑑,𝑑]: Δ⃑𝐻=π‘šο„Έβƒ‘π‘Žπ‘‘οοοŽ‘οŽ d or, in scalar form, Δ𝐻=π‘šο„Έπ‘Žπ‘‘.d

Let us look at an example of how this can be used to find a change in momentum.

Example 6: Finding the Change in Momentum of a Body given Its Acceleration

A body of mass 5 kg moves along a straight line. At time 𝑑 seconds, its acceleration is given by π‘Ž=(βˆ’6π‘‘βˆ’8)/ms. Find the change in momentum in the time interval 6≀𝑑≀9.

Answer

We have been given the acceleration, π‘Ž, of a body in terms of time, 𝑑, and also as a scalar. Therefore, in order to find its change in momentum, we can use the following formula: Δ𝐻=π‘šο„Έπ‘Žπ‘‘.d

From the question, we can see that we have π‘š=5kg, π‘Ž=(βˆ’6π‘‘βˆ’8)/ms, 𝑑=6s, and 𝑑=9s. Substituting these values into our formula, we have Δ𝐻=5ο„Έ(βˆ’6π‘‘βˆ’8)𝑑.d

Using the power rule for integration, we increase the powers of 𝑑 by 1 and then divide each term by the new power of 𝑑. This gives us Δ𝐻=5ο‘ο€Ήβˆ’3π‘‘βˆ’8𝑑.

Now, we just need to substitute in our values for 𝑑 and simplify as follows: Δ𝐻=5ο€Ήο€Ήβˆ’3Γ—9βˆ’8Γ—9ο…βˆ’ο€Ήβˆ’3Γ—6βˆ’8Γ—6=5((βˆ’243βˆ’72)βˆ’(βˆ’54βˆ’48))=βˆ’795.

We must not forget to include the units, which gives us our solution of Δ𝐻=βˆ’795β‹…/.kgms

Key Points

  • The momentum, ⃑𝐻, of an object is the product of its mass, π‘š, and its velocity, ⃑𝑣: ⃑𝐻=π‘šβƒ‘π‘£.
  • Momentum is typically measured in units of kilogram-metres per second, or kgβ‹…m/s.
  • Sometimes, we may have to use the kinematic equations to find the velocity of an object in order to calculate its momentum.
  • If we are given a function for the position of an object at a time 𝑑, we can take the derivative of that function with respect to time to get a function for the velocity and then use that function in calculating the momentum.

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